
chain rule problem
Find the derivative of the function. y = 2^3  x^2
cant get the right answer, i think its partly due to me not knowing the derivative of 2^x.
I did:
outside function = 2^x
outside deriv.= 2^xlog(2) <this is what i got from an online deriv. calc
inside = 3x^2
inside deriv = 2x


Please post the original function in a clear manner.


I dont undersand the whole inside and outside thing you got going on, but i know that since it is an additive functionjust take the derivative of each term. Like you did in the inside part and put them together.
The answer would be:
y'= 3x^22x
which folows f(x)= x^(n)
nx^(n1)

Quote:
Originally Posted by
Evan.Kimia
oh sorry i get your question now :)

Quote:
Originally Posted by
Evan.Kimia
Take y=2^(3x^2)
youre probably making a simple mistake
16*log(2)*x/2^x^2

$\displaystyle y= 2^{3x^2}$
make $\displaystyle u = 3x^2 \Rightarrow \frac{du}{dx} = 2x$
and $\displaystyle y= 2^u \Rightarrow \frac{dy}{du} = \ln(2)2^u$
$\displaystyle \frac{dy}{dx} = \frac{dy}{du}\times \frac{du}{dx}$
$\displaystyle \frac{dy}{dx} = \ln(2)2^u\times 2x$
$\displaystyle \frac{dy}{dx} = \ln(2)2^{3x^2}\times 2x$
$\displaystyle \frac{dy}{dx} = 2\ln(2)x2^{3x^2} $